document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1079291>Question 1079291</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve this\r<BR>\n" );
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document.write( "Suppose the quantity Q, in mg, of medicine in a patient’s bloodstream is decreasing by 25% each 40 minutes. Let us write Q = f(T), where T is in minutes since the injection of medicine. Suppose the injection contains 300 mg of medicine.<BR>\n" );
document.write( "(a) Construct a table of values for Q as a function of T over a few hours.<BR>\n" );
document.write( "(b) Find a formula for f in the form A × B^T, or an equivalent form.<BR>\n" );
document.write( "(c) Let th be the time taken for the quantity of medicine in the bloodstream to halve. By a process of guessing and checking, estimate the value of Th (accurate to 3 significant figures).<BR>\n" );
document.write( "(d) Determine the exact value of Th.<BR>\n" );
document.write( "(e) Show that the halving time is constant. That is, show that starting from an arbitrary time<BR>\n" );
document.write( "T= a, the quantity of medicine is halved after an additional time Th.<BR>\n" );
document.write( "(f) Sketch a graph of the function over a 4 hour period since the injection.<BR>\n" );
document.write( "(g) The patient can receive a new injection when the quantity of medicine is less than 2% of the original dose. For simplicity, this time delay is measured in a whole number of hours. Determine the wait time. For medicinal safety, should we round up or down to the nearest hour?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #693690 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=693690\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> f = p * (1 + r) ^ n\r<BR>\n" );
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document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
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document.write( "the quantity of medicine is decreasing by 25% every 40 minutes.\r<BR>\n" );
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document.write( "your time period is therefore every 40 minutes.\r<BR>\n" );
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document.write( "your rate is -25% every 40 minutes.\r<BR>\n" );
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document.write( "in the formula, you use the decimal equivalent of the percent which is equal to the percent / 100.\r<BR>\n" );
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document.write( "the interest rate per time period is therefore -.25\r<BR>\n" );
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document.write( "the formula of f = p * (1 + r) ^ n becomes\r<BR>\n" );
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document.write( "f = p * (1 - .25) ^ n which becomes:\r<BR>\n" );
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document.write( "f = p * .75 ^ n\r<BR>\n" );
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document.write( "in order for the value of the medicine to be equal to 1/2 what it was, the formula becomes:\r<BR>\n" );
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document.write( "1/2 = .75^n \r<BR>\n" );
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document.write( "this is because f is 1/2 and p is 1, and 1 * .75 ^ n is the same as .75 ^ n.\r<BR>\n" );
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document.write( "you would solve for n using logs.\r<BR>\n" );
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document.write( "take the log of both sides of the eqution to get log(1/2) = log(.75^n)\r<BR>\n" );
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document.write( "because log(b^n) = n*log(b), this becomes:\r<BR>\n" );
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document.write( "log(1/2) = n*log(.75)\r<BR>\n" );
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document.write( "divide both sides of this equation by log(.75) and you get:\r<BR>\n" );
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document.write( "log(1/2) / log(.75) = n\r<BR>\n" );
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document.write( "solve for n to get n = 2.40942084\r<BR>\n" );
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document.write( "this means that that the original amount will be half of what it was in 2.40942084 time periods that are 40 minutes each.\r<BR>\n" );
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document.write( "translate this to minutes, and it becomes n = 2.40942084 * 40 = 96.37683359 minutes.\r<BR>\n" );
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document.write( "this formula will work regardless of the value of the milligrams in the blood stream.\r<BR>\n" );
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document.write( "start with 300 mg\r<BR>\n" );
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document.write( "you will have 150 mg in the blood stream in 2.40942084 time periods.\r<BR>\n" );
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document.write( "f = 300 * .75 ^ 2.40942084 = 150\r<BR>\n" );
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document.write( "f = 150 * .75 ^ 2.40942084 = 75\r<BR>\n" );
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document.write( "f = 75 * .75 ^ 2.40942084 = 37.5\r<BR>\n" );
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document.write( "etc.\r<BR>\n" );
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document.write( "the amount of drug in the bloodstream will be half of what it was every 2.40942084 time periods.\r<BR>\n" );
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document.write( "your original table can be set up in 40 minute intervals.\r<BR>\n" );
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document.write( "when n = 0, the value in this table will be 300 * .75 ^ 0 = 300.\r<BR>\n" );
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document.write( "when n = 1, the value in this table will be 300 * .75 ^ 1 = 225\r<BR>\n" );
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document.write( "when n = 2, the value in this table will be 300 * .75 ^ 2 = 168.75\r<BR>\n" );
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document.write( "when n = 3, the value in this table will be 300 * .75 ^ 3 = 126.5625\r<BR>\n" );
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document.write( "etc.\r<BR>\n" );
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document.write( "for the remaining dose being 2% of the original dose, you would calculate the time required in 40 minute intervals as follows:\r<BR>\n" );
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document.write( "2% of 300 = .02 * 300 = 6 mg.\r<BR>\n" );
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document.write( "in the formula of f = p * .75 ^ n, you would get:\r<BR>\n" );
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document.write( "6 = 300 * .75 ^ n\r<BR>\n" );
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document.write( "you will want to solve for n in this equation.\r<BR>\n" );
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document.write( "divide both sides of the equation by 300 to get:\r<BR>\n" );
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document.write( "6/300 = .75 ^ n\r<BR>\n" );
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document.write( "take the log of both sides of this equation to get:\r<BR>\n" );
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document.write( "log(6/300) = log(.75^n)\r<BR>\n" );
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document.write( "this becomes log(6/300) = n * log(.75)\r<BR>\n" );
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document.write( "divide both sides of the equation by log(.75) to get:\r<BR>\n" );
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document.write( "log(6/300) / log(.75) = n\r<BR>\n" );
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document.write( "solve for n to get n = 13.59842472\r<BR>\n" );
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document.write( "the original dose of 300 mg should reduce to 6 mg in 13.59842472 time periods that are 40 minutes each.\r<BR>\n" );
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document.write( "that's equivalent to 13.59842472 * 40 = 543.9369888 minutes which is equivalent to 543.9369888 / 60 = 9.06561648 hours.\r<BR>\n" );
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document.write( "to be on the safe side, you would round down so there is at least 2% of the medicine in the bloodstream.\r<BR>\n" );
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document.write( "to graph the equation, replace f with y and replace n with x to get:\r<BR>\n" );
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document.write( "y = 300 * .75^x\r<BR>\n" );
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document.write( "your graph will look  like this:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2017/050311.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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document.write( "if your want your time intervals to be minutes instead of 40 minutes, you would do the following.\r<BR>\n" );
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document.write( ".75 ^ 1/40 = .9928337491\r<BR>\n" );
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document.write( "your growth rate per minute would be .9928337491 per 1 minute interval.\r<BR>\n" );
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document.write( "your time interval is now in minutes.\r<BR>\n" );
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document.write( "i will graph the equation of y = 300 * .9928337491 ^ x to show you that this is equivalent to the same equation when the time period is in 40 minute intervals.\r<BR>\n" );
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document.write( "your graph looks like this.\r<BR>\n" );
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document.write( "the value of x is now every minute rather than every 40 minutes.\r<BR>\n" );
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document.write( "you can also translate to hours.\r<BR>\n" );
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document.write( "your do that as follows:\r<BR>\n" );
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document.write( "hours relative to 40 minutes are 60/40 = 6/4 = 3/2\r<BR>\n" );
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document.write( ".75 ^ (3/2) = .6495190528\r<BR>\n" );
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document.write( "your equation becomes y = 300 * .6495190528 ^ x\r<BR>\n" );
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document.write( "the value of x is now every hour instead of every 40 minute interval.\r<BR>\n" );
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document.write( "the graph of that equation looks like this:\r<BR>\n" );
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document.write( "if you want to sketch a 4 hour graph, it would look like this:\r<BR>\n" );
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